Example 10 - Chapter 8 Class 11 Binomial Theorem (Deleted)
Last updated at Sept. 3, 2021 by Teachoo
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Examples
Example 10 Find the term independent of x in the expansion of (3/2 π₯^2 " β " 1/3π₯)^6,x > 0. Calculating general term We know that general term of expansion (a + b)n is Tr + 1 = nCr (a)nβr.(b)n For general term of expansion (3/2 π₯^2 " β " 1/3π₯)^6 Putting n = 6 , a = 3/2 π₯^2 , b = "β" 1/3π₯ Tr + 1 = 6Cr (π/π π^π )^(π β π) ((βπ)/ππ)^π = 6Cr (3/2)^(6 β π) (π₯^2 ")" ^((6 β π)) ((β1)/3 "Γ" 1/π₯)^π = 6Cr (3/2)^(6 β π) (π₯")" ^(2(6 β π)) ((β1)/3)^π (1/π₯)^π = 6Cr (3/2)^(6 β π) (π₯")" ^(12 β 2π) ((β1)/3)^π (π₯)^(βπ) = 6Cr (3/2)^(6 β π) ((β1)/3)^π (π₯")" ^(12 β 2π) (π₯)^(βπ) = 6Cr (3/2)^(6 β π) ((β1)/3)^π (π₯")" ^(12 β 2π β π) = 6Cr (3/2)^(6 β π) ((β1)/3)^π (π₯")" ^(12 β3π) We need to find the term independent of x So, power of x is 0 π₯^(12 β 3π) = x0 Comparing powers 12 β 3r = 0 12 = 3r 12/3 = r 4 = r r = 4 Putting r = 4 in (1) T4+1 = 6C4 (3/2)^(6 β 4) ((β1)/3)^4 (π₯")" ^(12 β3(4)) T5 = 6C4 (3/2)^2 (1/3^4 ) (π₯")" ^(12 β12) = 6C4 (3^2/2^2 )(1/3^4 ) (π₯")" ^0 = 6C4 (1/2^2 )(3^2/3^4 ) (1) = 6C4 (1/2^2 )(1/3^2 ) = 6!/4!(6 β 4)! (1/4)(1/9) = 6!/4!(2)! (1/4)(1/9) = (6(5)(4)!)/4!(2)! (1/4)(1/9) = (5 )/12 Hence, the term which is independent of x is 5th term = T5 = π/ππ